CHAPTER 9 CONTACT EQUILIBRIUM PROCESSES

Transcription

1 CHAPTER 9 CONTACT EQUILIBRIUM PROCESSES Biological raw materials are usually mixtures, and to prepare foods it may be necessary to separate some of the components of the mixtures. One method, by which this separation can be carried out, is by the introduction of a new phase to the system and allowing the components of the original raw material to distribute themselves between the phases. For example, freshly dug vegetables have another phase, water, added to remove unwanted earth; a mixture of alcohol and water is heated to produce another phase, vapour, which is richer in alcohol than the mixture. By choosing the conditions, one phase is enriched whilst the other is depleted in some component or components. The maximum separation is reached at the equilibrium distribution of the components, but in practice separation may fall short of this as equilibrium is not attained. The components are distributed between the phases in accordance with equilibrium distribution coefficients, which give the relative concentrations in each phase when equilibrium has been reached. The two phases can then be separated by simple physical methods such as gravity settling. This process of contact, redistribution, and separation gives the name contact equilibrium separations. Successive stages can be used to enhance the separation. An example is in the extraction of edible oil from soya beans. Beans containing oil are crushed, and then mixed with a solvent in which the oil, but not the other components of the beans, is soluble. Initially, the oil will be distributed between the beans and the solvent, but after efficient crushing and mixing the oil will be dissolved in the solvent. In the separation, some solvent and oil will be retained by the mass of beans; these will constitute one stream and the bulk of the solvent and oil the other. This process of contacting the two streams, of crushed beans and solvent, makes up one contact stage. To extract more oil from the beans, further contact stages can be provided by mixing the extracted beans with a fresh stream of solvent. For economy and convenience, the solvent and oil stream from another extraction is often used instead of fresh solvent. So two streams, one containing beans and the other starting off as pure solvent, can move counter current to each other through a series of contact stages with progressive contacting followed by draining. In each stage of the process in which the streams come into contact, the material being transferred is distributed in equilibrium between the two streams. By removing the streams from the contact stage and contacting each with material of different composition, new equilibrium conditions are established and so separation can proceed. In order to effect the desired separation of oil from beans, the process itself has introduced a further separation problem - the separation of the oil from the solvent. However, the solvent is chosen so that this subsequent separation is simple, for example by distillation. In some cases, such as washing, further separation of dissolved material from wash water may not be necessary and one stream may be rejected as waste. In other cases, such as distillation, the two streams are generated from the mixture of original components by vaporization of part of the mixture.

2 The two features that are common to all equilibrium contact processes are the attainment of, or approach to, equilibrium and the provision of contact stages. Equilibrium is reached when a component is so distributed between the two streams that there is no tendency for its concentration in either stream to change. Attainment of equilibrium may take appreciable time, and only if this time is available will effective equilibrium be reached. The opportunity to reach equilibrium is provided in each stage, and so with one or more stages the concentration of the transferred component changes progressively from one stream to the other, providing the desired separation. Some examples of contact equilibrium separation processes are: Gas absorption Extraction and washing Crystallization Membrane separations Distillation In addition, drying and humidification, and evaporation can be considered under this general heading for some purposes, but it seemed more appropriate in this book to take them separately. For the analysis of these processes, there are two major sets of quantitative relationships; the equilibrium conditions that determine how the components are distributed between the phases, and the material flow balances that follow the progression of the components stage by stage. PART 1. THEORY CONCENTRATIONS The driving force, which produces equilibrium distributions, is considered to be proportional at any time to the difference between the actual concentration and equilibrium concentration of the component being separated.. Thus, concentrations in contact equilibrium separation processes are linked with the general driving force concept. Consider a case in which initially all of the molecules of some component A of a gas mixture are confined by a partition in one region of a system. The partition is then removed. Random movement among the gas molecules will, in time, distribute component A through the mixture. The greater the concentration of A in the partitioned region, the more rapidly will diffusion occur across the boundary once the partition is removed. The relative proportions of the components in a mixture or a solution are expressed in terms of the concentrations. Any convenient units may be used for concentration, such as gg -1,gkg -1, gg -1, percentages, parts per million, and so on. Because the gas laws are based on numbers of molecules, it is often convenient to express concentrations in terms of the relative numbers of molecules of the components. The unit in

3 this case is called the molecular fraction, shortened to mole fraction, which has been introduced in Chapter 2. The mole fraction of a component in a mixture is the proportion of the number of molecules of the component present to the total number of the molecules of all the components. In a mixture which contains wa kg of component A of molecular weight MA and wb kg of component B of molecular weight MB, the mole fraction: xa = number of moles of A number of moles of A + number of moles of B xa = (wa /MA )_/ (wa /MA + wb /MB) (9.1) xb = (wb /MB ) / (wa /MA + wb/mb ) (9.2) Notice that (xa + xb) = 1, and so, xb = (1 - xa) The definition of the mole fraction can be extended to any number of components in a multicomponent mixture. The mole fraction of any one component again expresses the relative number of molecules of that component, to the total number of molecules of all the components in the mixture. Exactly the same method is followed if the weights of the components are expressed in grams. The mole fraction is a ratio, and so has no dimensions. EXAMPLE 9.1. Mole fractions of ethanol in water A solution of ethanol in water contains 30% of ethanol by weight. Calculate the mole fractions of ethanol and water in the solution. Molecular weight of ethanol, C2H5OH, is 46 and the molecular weight of water, H2O, is 18. Since, in 100 kg of the mixture there are 30 kg of ethanol and 70 kg of water, mole fraction of ethanol = (30/46)/ (30/ /18) = mole fraction of water = (70/18)/ (30/ /18) = = ( ) Concentrations of the components in gas mixtures can be expressed as weight fractions, mole fractions, and so on. When expressed as mole fractions, they can be related to the partial pressure of the components. The partial pressure of a component is that pressure which the component would exert if it alone occupied the whole volume of the mixture. Partial pressures of the components are additive, and their sum is equal to the total pressure of the mixture. The partial pressures and the mole fractions are proportional, so that the total pressure is made up from the sum of all the partial pressures, which are in the ratios of the mole fractions of the components. If a gas mixture exists under a total pressure P and the mixture comprises a mole fraction xa of component A, a mole fraction xb of component B, a mole fraction xc of component C and so on, then

4 P = PxA + PxB + PxC +.. = pa + pb + pc +.. (9.3) where pa, pb, pc, are the partial pressures of components A, B, C... In the case of gas mixtures, it is also possible to relate weight and volume proportions, as Avogadro's Law states that under equal conditions of temperature and pressure, equal volumes of gases contain equal numbers of molecules. This can be put in another way by saying that in a gas mixture, volume fractions will be proportional to mole fractions. EXAMPLE 9.2. Mole fractions in air Air is reported to contain 79 parts of nitrogen to 21 parts of oxygen, by volume. Calculate the mole fraction of oxygen and of nitrogen in the mixture and also the weight fractions and the mean molecular weight. Molecular weight of nitrogen is 28, and of oxygen 32. Since mole fractions are proportional to volume fractions, mole fraction of nitrogen = 79/( ) = 0.79 mole fraction of oxygen = 21/( ) = 0.21 The molecular weight of nitrogen, N2, is 28 and of oxygen, O2, is 32. The weight fraction of nitrogen is given by: weight of nitrogen/ (weight of nitrogen + weight of oxygen)= (79 x 28)/(79 x x 32) = 0.77 Similarly the weight fraction of oxygen = (21 x 32)/ (79 x x 32) = As the sum of the two weight fractions must add to 1, the weight fraction of the oxygen could have been found by the subtraction of (1-0.77) = To find the mean molecular weight, we must find the weight of one mole of the gas: 0.79 moles of N2 weighing 0.79 x 28 kg = 22.1 kg plus 0.21 moles of O2 weighing 0.21 x 32kg = 6.7 kg make up 1 mole of air weighing = 28.8 kg and so Mean molecular weight of air is 28.8, say 29. GAS/LIQUID EQUILIBRIA Molecules of the components in a liquid mixture or a solution have a tendency to escape from the liquid surface into the gas above the solution. The escaping tendency sets up a pressure above the surface of the liquid, owing to the resultant concentration of the escaped molecules. This pressure is called the vapour pressure of the liquid.

5 The magnitude of the vapour pressure depends upon the liquid composition and upon the temperature. For pure liquids, vapour/pressure relationships have been tabulated and may be found in reference works such as Perry (1997) or the International Critical Tables. For a solution or a mixture, the various components in the liquid each exert their own partial vapour pressures. When the liquid contains various components it has been found that, in many cases, the partial vapour pressure of any component is proportional to the mole fraction of that component in the liquid. That is, pa = HA xa (9.4) where pa is the partial vapour pressure of component A, HA is a constant for component A at a given temperature and xa is the mole fraction of component A in the liquid. This relationship is approximately true for most systems and it is known as Henry's Law. The coefficient of proportionality HA is known as the Henry's Law constant and has units of kpa mole fraction -1. In reverse, Henry's Law can be used to predict the solubility of a gas in a liquid. If a gas exerts a given partial pressure above a liquid, then it will dissolve in the liquid until Henry's Law is satisfied and the mole fraction of the dissolved gas in the liquid is equal to the value appropriate to the partial pressure of that gas above the liquid. The reverse prediction can be useful for predicting the gas solubility in equilibrium below imposed gaseous atmospheres of various compositions and pressures. EXAMPLE 9.3. Solubility of carbon dioxide in water Given that the Henry's Law constant for carbon dioxide in water at 25 o C is 1.6x10 5 kpa mole fraction -1, calculate the percentage solubility by weight of carbon dioxide in water under these conditions and at a partial pressure of carbon dioxide of 200kPa above the water. From Henry's Law p = H x 200 = 1.6 x 10 5 x But since ( wh20/18)» (wco2/44) x = = 1.25x10-3 = (wco2 /44)/( wh20/18+ wco2 /44) 1.25 x 10-3 (wco2/44) / ( wh20/18) and so (wco2/ wh20) 1.25 x 10-3 / (44/18) = 3.1 x 10-3 = 3.1 x 10-1 % = 0.31%

6 SOLID/LIQUID EQUILIBRIA Liquids have a capacity to dissolve solids up to an extent, which is determined by the solubility of the particular solid material in that liquid. Solubility is a function of temperature and, in most cases, solubility increases with rising temperature. A solubility curve can be drawn to show this relationship, based on the equilibrium concentration in solution measured in convenient units, for example kg100kg -1 water as a function of temperature. Such a curve is illustrated in Fig. 9.1 for sodium nitrite in water. Figure 9.1 Solubility of sodium nitrite in water There are some relatively rare systems in which solubility decreases with temperature, and they provide what is termed a reversed solubility curve. The equilibrium solution, which is ultimately reached between solute and solvent, is called a saturated solution, implying that no further solute can be taken into solution at that particular temperature. An unsaturated solution plus solid solute is not in equilibrium, as the solvent can dissolve more of the solid. When a saturated solution is heated, if it has a normal solubility curve the solution then has a capacity to take up further solute material, and so it becomes unsaturated. Conversely, when a saturated solution is cooled it becomes supersaturated, and at equilibrium that solute which is in excess of the solubility level at the particular temperature will come out of solution, normally as crystals. However, this may not occur quickly and in the interim the solution is spoken of as supersaturated and it is said to be in a metastable state.

7 EQUILIBRIUM / CONCENTRATION RELATIONSHIPS A contact equilibrium separation process is designed to reduce the concentration of a component in one phase, or flowing stream in a continuous process, and increase it in another. Conventionally, and just to distinguish one stream from another, one is called the overflow and the other the underflow. The terms referred originally to a system of two immiscible liquids, one lighter (the overflow) and the other heavier (the underflow) than the other, and between which the particular component was transferred. When there are several stages, the overflow and underflow streams then move off in opposite directions in a counter flow system. Following standard chemical engineering nomenclature, the concentration of the component of interest in the lighter stream, that is the stream with the lower density, is denoted by y. For example, in a gas absorption system, the light stream would be the gas; in a distillation column, it would be the vapour stream; in a liquid extraction system, it would be the liquid overflow. The concentration of the component in the heavier stream is denoted by x. Thus we have two streams; in one the concentration of the component is y and in the other, the heavier stream, it is x. For a given system, it is often convenient to plot corresponding (equilibrium) values of y against x in an equilibrium diagram. In the simple case of multistage oil extraction with a solvent, equilibrium is generally attained in each stage. The concentration of the oil is the same in the liquid solution spilling over or draining off in the overflow as it is in the liquid in the underflow containing the solids, so that in this case y = x and the equilibrium/concentration diagram is a straight line. In gas absorption, such relationships as Henry's Law relate the concentration in the light gas phase to that in the heavy liquid phase and so enable the equilibrium diagram to be plotted. In crystallization, the equilibrium concentration corresponds to the solubility of the solute at the particular temperature. Across a membrane, there is some equilibrium distribution of the particular component of interest. If the concentration in one stream is known, the equilibrium diagram allows us to read off the corresponding concentration in the other stream if equilibrium has been attained. The attainment of equilibrium takes time and this has to be taken into account when considering contact stages. The usual type of rate equation applies, in which the rate is given by the driving force divided by a resistance term. The driving force is the extent of the departure from equilibrium and generally is measured by concentration differences. Resistances have been classified in different ways but they are generally assumed to be concentrated at the phase boundary. Stage contact systems in which equilibrium has not been attained are beyond the scope of this book. In many practical cases, allowance can be made for non-attained equilibrium by assuming an efficiency for each stage, in effect a percentage of equilibrium actually attained.

8 OPERATING CONDITIONS In a series of contact stages, in which the components counter flow from one stage to another, mass balances can be written around any stage, or any number of stages. This enables operating equations to be set down to connect the flow rates and the compositions of the streams. Consider the generalized system shown in Fig. 9.2, in which there is a stage contact process operating with a number of stages and two contacting streams. (a) (b) Figure 9.2 Contact equilibrium stages In Fig. 9.2(a), the mass flow of the light stream is denoted by V and the flow of the heavy stream by L and the concentration in the light phase by y and in the heavy phase by x. Taking a balance over the first n stages as in Fig. 9.2(b), we can write, for the total flow, mass entering must equal mass leaving, so: Vn+1 + La = Va + Ln and for the component being exchanged: Vn+ 1yn+ 1 + Laxa = Vaya + Ln xn where V is the mass flow rate of the light stream, L is the flow rate of the heavy stream, y is the concentration of the component being exchanged in the light stream and x is the concentration of the component being exchanged in the heavy stream. In the case of the subscripts, n denotes conditions at equilibrium in the nth stage, (n + 1) denotes conditions at equilibrium in the (n + 1)th stage and a denotes the conditions of the streams entering and leaving stage 1, one being raw material and one product from that stage Eliminating Vn+1 between these equations, we have: and so, Vn+1 = Ln - La + Va (Ln - La+ Va ) yn+1 = Vaya + Ln xn - Laxa yn+1 = (Vaya + Ln xn - Laxa )/(Ln - La + Va ) yn+1 = xn [Ln / (Ln - La + Va )] + [(Vaya - Laxa )/ (Ln - La + Va )] (9.5) This is an important equation as it expresses the concentration in one stream (the lighter stream) in the (n + 1)th stage in terms of the concentration in the other (heavier)stream in the nth stage.

9 In many practical cases in which equal quantities, or equal molal quantities, of the carrying streams move from one stage to another, that is where the flow rates are the same in all contact stages, then for: heavier phase Ln+1 = Ln =... La = L lighter phase Vn+1 = Vn = Va. = V A simplified equation can be written for such cases: yn+1 = xn L / V + ya - xa L / V (9.6) CALCULATION OF SEPARATION IN CONTACT EQUILIBRIUM PROCESSES The separation, which will be effected in a given series of contact stages, can be calculated by combining the equilibrium and the operating relationships. Starting at one end of the process, the terminal separation can be calculated from the given set of conditions. Knowing, say, the x value in the first stage, x1, the equilibrium condition gives the corresponding value of y in this stage, y1 Then eqn. (9.5) or eqn. (9.6) can be used to obtain y2 then the equilibrium conditions give the corresponding x2, and so on... EXAMPLE 9.4. Single stage steam stripping, of taints from cream A continuous deodorizing system, involving a single stage steam stripping operation, is under consideration for the removal of taints from cream. If the taint component is present in the cream to the extent of 8 parts per million (ppm) and if steam is to be passed through the contact stage in the proportions of 0.75kg steam to every 1kg cream, calculate the concentration of the taint in the leaving cream. The equilibrium concentration distribution of the taint has been found experimentally to be in the ratio of 1 in the cream to 10 in the steam and it is assumed that equilibrium is reached in each stage. Call the concentration of the taint in the cream x, and in the steam y, both as mass fractions, From the condition that, at equilibrium, the concentration of the taint in the steam is 10 times that in the cream: 10x = y and in particular, 10x1 = y1 Now, y1 the concentration of taint in the steam leaving the stage is also the concentration in the output steam: y1 = ya = 10x1 The incoming steam concentration = y2 = yb = 0 as there is no taint in the entering steam. The taint concentration in the entering cream is xa = 8ppm.

10 These are shown diagrammatically in Fig Steam Figure 9.3 Flows into and out from a stage Cream The problem is to determine x1 the concentration of taint in the product cream. Basis is 1kg of cream. The mass ratio of stream flows is 1 of cream to 0.75 of steam and if no steam is condensed this ratio will be preserved through the stage. 1/0.75 = 1.33 is the ratio of cream flow rate to steam flow rate = L/V. Applying eqn. (9.6) to the one stage n = 1, y2 = xn L/V + ya - xa L/V y2 = 0 = x x1-8 x 1.33 xl = 10.64/11.33 = 0.94ppm which is the concentration of the taint in the leaving cream, having been reduced from 8 ppm. This simple example could have been solved directly without using the formula, but it shows the way in which the formula and the equilibrium conditions can be applied. McCabe -Thiele plot Based on the step-by-step method of calculation, it was suggested by McCabe and Thiele (1925) that the operating and equilibrium relationships could very conveniently be combined in a single graph called a McCabe -Thiele plot. The essential feature of their method is that whereas the equilibrium line is plotted directly, xn against yn, the operating relationships are plotted as xn against yn+1. Inspection of eqn. (9.5) shows that it gives yn+1 in terms of xn and the graph of this is called the operating line. In the special case of eqn. (9.6), the operating line is a straight line whose slope is L/V and whose intercept on the y-axis is (ya - xa L/V).

11 Considering any stage in the process, it might be for example the first stage, we have the value of y from given or overall conditions. Proceeding at constant y to the equilibrium line we can then read off the corresponding value of x, which is x1. From x1 we proceed at constant x across to the operating line at which the intercept gives the value of y2. Then the process can be repeated for y2 to x2, then to y3, and so on. Drawing horizontal and vertical lines to show this, as in Fig. 9.4 in Example 9.5, a step pattern is traced out on the graph. Each step represents a stage in the process at which contact is provided between the streams, and the equilibrium attained. Proceeding step-by-step, it is simple to insert sufficient steps to move to a required final concentration in one of the streams and so to be able to count the number of stages of contact needed to obtain this required separation. PART 2 APPLICATIONS GAS ABSORPTION Gas absorption/desorption is a process in which a gaseous mixture is brought into contact with a liquid and during this contact a component is transferred between the gas stream and the liquid stream. The gas may be bubbled through the liquid, or it may pass over streams of the liquid, arranged to provide a large surface through which the mass transfer can occur. The liquid film in this latter case can flow down the sides of columns or over packing, or it can cascade from one tray to another with the liquid falling and the gas rising in the counter flow. The gas, or components of it, either dissolves in the liquid (absorption)or extracts a volatile component from the liquid (desorption). An example of absorption is found in hydrogenation of oils, in which the hydrogen gas is bubbled through the oil with which it reacts. Generally, there is a catalyst present also to promote the reaction. The hydrogen is absorbed into the oil, reacting with the unsaturated bonds in the oil to harden it. Another example of gas absorption is in the carbonation of beverages. Carbon dioxide under pressure is dissolved in the liquid beverage, so that when the pressure is subsequently released on opening the container, effervescence occurs. An example of desorption is found in the steam stripping of fats and oils in which steam is brought into contact with the liquid fat or oil, and undesired components of the fat or oil pass out with the steam. This is used in the deodorizing of natural oils before blending them into food products such as margarine, and in the stripping of unwanted flavours from cream before it is made into butter. The equilibrium conditions arise from the balance of concentrations of the gas or the volatile flavour, between the gas and the liquid streams. In the gas absorption process, sufficient time must be allowed for equilibrium to be attained so that the greatest possible transfer can occur and, also, opportunity must be provided for contacts between the streams to occur under favourable conditions.

12 Rate of Gas Absorption The rates of mass transfer in gas absorption are controlled by the extent of the departure of the system from the equilibrium concentrations and by the resistance offered to the mass transfer by the streams of liquid and gas. Thus, we have the familiar expression: rate of absorption = driving force/resistance, The driving force is the extent of the difference between the actual concentrations and the equilibrium concentrations. This is represented in terms of concentration differences. For the resistance, the situation is complicated, but for practical purposes it is adequate to consider the whole of the resistance to be concentrated at the interface between the two streams. At the interface, two films are postulated, one in the liquid and one in the gas. The two-film theory of Lewis and Whitman defines these resistances separately, a gas film resistance and a liquid film resistance. They are treated very similarly to surface heat transfer coefficients and the resistances of the two films can be combined in an overall resistance similar to an overall heat transfer coefficient. The driving forces through each of the films are considered to be the concentration differences between the material in the bulk liquid or gas and the material in the liquid or gas at the interface. In practice, it is seldom possible to measure interfacial conditions and overall coefficients are used giving the equation dw/dt = KlA (x* - x) = KgA(y - y*) (9.7) where dw/dt is the quantity of gas passing across the interface in unit time, Kl is the overall liquid mass transfer coefficient, Kg is the overall gas mass transfer coefficient, A is the interfacial area and x, y are the concentrations of the gas being transferred, in the liquid and gas streams respectively. The quantities of x* and y* are introduced into the equation because it is usual to express concentrations in the liquid and in the gas in different units. With these, x* represents the concentration in the liquid which would be in equilibrium with a concentration y in the gas and y* the concentration in the gas stream which would be in equilibrium with a concentration x in the liquid. Equation (9.7) can be integrated and used to calculate overall rates of gas absorption or desorption. For details of the procedure, reference should be made to works such as Perry (1997), Charm (1971), Coulson and Richardson (1978) or McCabe and Smith (1975). Stage Equi1ibrium Gas Absorption The performance of counter current, stage contact, gas absorption equipment can be calculated if the operating and equilibrium conditions are known. The liquid stream and the gas stream are brought into contact in each stage and it is assumed that sufficient time is allowed for equilibrium to be reached. In cases where sufficient time is not available for equilibration, the rate equations have to be introduced and this complicates the analysis. However, in many cases of practical importance in the food industry, either the time is sufficient to reach equilibrium, or else the calculation can be carried out on the assumption that it is and a stage

13 efficiency term, a fractional attainment of equilibrium, introduced to allow for the conditions actually attained. Appropriate efficiency values can sometimes be found from published information, or sought experimentally. After the streams in a contact stage have come to equilibrium, they are separated and then pass in opposite directions to the adjacent stages. The separation of the gas and the liquid does not generally present great difficulty and some form of cyclone separator is often used. In order to calculate the equipment performance, operating conditions must be known or found from the mass balances. Very often the known factors are: gas and liquid rates of flow, inlet conditions of gas and liquid, one of the outlet conditions, and equilibrium relationships. The processing problem is to find how many contact stages are necessary to achieve the concentration change that is required. An overall mass balance will give the remaining outlet condition and then the operating line can be drawn. The equilibrium line is then plotted on the same diagram, and the McCabe-Thiele construction applied to solve the problem. EXAMPLE 9.5. Multiple stage steam stripping of taints from cream In Example 9.4, a calculation was made for a single stage, steam stripping process to remove taints in the cream, by contact with a counter flow current of steam. Consider, now, the case of a rather more difficult taint to remove in which the equilibrium concentration of the taint in the steam is only 7.5 times as great as that in the cream. If the relative flow rates of cream and steam are given in the ratio 1: 0.75, how many contact stages would be required to reduce the taint concentration in the cream to 0.3ppm assuming (a) 100% stage efficiency and (b) 70% stage efficiency? The initial concentration of the taint is 10ppm. Mass balance Inlet cream taint concentration = 10ppm = xa Outlet cream taint concentration = 0.3ppm Inlet steam taint concentration = xn = 0ppm = yn+1 Assume a cream flow rate L = 100 arbitrary units so steam flow rate V = 75 If y1 is the outlet steam taint concentration, total taint into equipment = total taint out of equipment. 100(10) = 75y (0.3) 100(10-0.3) = 75y1 Therefore y1 = 12.9ppm = ya From eqn. (9.6) Yn+1 = xn L/V + ya - xa L/V

14 Yn+1 = xn (100/75) (100/75) = 1.33 xn Equilibrium condition: Yn = 7.5xn Figure 9.4 Steam stripping of cream: McCabe -Thiele plot On the graph of Fig. 9.4 are shown the operating line, plotting xn against yn+1, and the equilibrium line in which xn, is plotted against yn. Starting from one terminal condition on the operating line, the stage contact steps are drawn in until the desired other terminal concentrations are reached. Each of the numbered horizontal lines represents one stage From the operating and equilibrium plotted on Fig. 9.4, it can be seen that two contact stages are sufficient to effect the required separation. The construction assumes 100% efficiency so that, with a stage efficiency of 70%, the number of stages required would be 2(100/70) and this equals approximately three stages. So the number of contact stages required assuming: (a) 100% efficiency = 2, and (b) 70% efficiency = 3. Notice that only a small number of stages is required for this operation, as the equilibrium condition is quite well removed from unity and the steam flow is of the same order as that of the cream. A smaller equilibrium constant, or a smaller relative steam-flow rate, would increase the required number of contact stages.

15 Gas Absorption Equipment Gas absorption equipment is designed to achieve the greatest practicable interfacial area between the gas and the liquid streams, so that liquid sprays and gas bubbling devices are often employed. In many cases, a vertical array of trays is so arranged that the liquid descends over a series of perforated trays, or flows down over ceramic packing that fills a tower. For the hydrogenation of oils, absorption is followed by reaction of the hydrogen with the oil, and a nickel catalyst is used to speed up the reactions. Also, pressure is applied to increase gas concentrations and therefore speed up the reaction rates. Practical problems are concerned with arranging distribution of the catalyst, as well as of oil and hydrogen. Some designs spray oil and catalyst into hydrogen, others bubble hydrogen through a continuous oil phase in which catalyst particles are suspended. For the stripping of volatile flavours and taints in deodorizing equipment, the steam phase is in general the continuous one and the liquid is sprayed into this and then separated. In one design of cream deodorizing plant, cream is sprayed into an atmosphere of steam and the two streams then pass on to the next stages, or the steam may be condensed and fresh steam used in the next stage. EXTRACTION AND WASHING It is often convenient to use a liquid in order to carry out a separation process. The liquid is thoroughly mixed with either solids or other liquid from which the component is to be removed and then the two streams are separated. In the case of solids, the separation of the two streams is generally by simple gravity settling. Sometimes it is the solution in the introduced liquid that is the product required, such as in the extraction of coffee from coffee beans with water. In other cases, the washed solid may be the product as in the washing of butter. The term washing is generally used where an unwanted constituent is removed in a stream of water. Extraction is also an essential stage in the sugar industry when soluble sucrose is removed by water extraction from sugar cane or sugar beet. Washing occurs so frequently as to need no specific examples. To separate liquid streams, the liquids must be immiscible, such as oil and water. Liquid/liquid extraction is the name used when both streams in the extraction are liquid. Examples of extraction are found in the edible oil industry in which oil is extracted from natural products such as peanuts, soya beans, rape seeds, sunflower seeds and so on. Liquid/liquid extraction is used in the extraction of fatty acids. Factors controlling the operation are: area of contact between the streams, time of contact, properties of the materials so far as the equilibrium distribution of the transferred component is concerned, number of contact stages employed.

16 In extraction from a solid, the solid matrix may hinder diffusion and so control the rate of extraction. Rate of Extraction The solution process can be considered in terms of the usual rate equation rate of solution = driving force/resistance. In this case, the driving force is the difference between the concentration of the component being transferred, the solute, at the solid interface and in the bulk of the solvent stream. For liquid/liquid extraction, a double film must be considered, at the interface and in the bulk of the other stream. For solution from a solid component, the equation can be written: dw/dt = Kl A(ys - y) (9.8) where dw/dt is the rate of solution, Kl is the mass transfer coefficient, A is the interfacial area, and ys and y, are the concentrations of the soluble component in the bulk of the liquid and at the interface. It is usually assumed that a saturated solution is formed at the interface and ys is the concentration of a saturated solution at the temperature of the system. Examination of eqn. (9.8) shows the effects of some of the factors, which can be used to speed up rates of solution. Fine divisions of the solid component increases the interfacial area A. Good mixing ensures that the local concentration is equal to the mean bulk concentration. In other words, it means that there are no local higher concentrations arising from bad stirring increasing the value of y and so cutting down the rate of solution. An increase in the temperature of the system will, in general, increase rates of solution by not only increasing Kl, which is related to diffusion, but also by increasing the solubility of the solute and so increasing ys. In the simple case of extraction from a solid in a contact stage, a mass balance on the solute gives the equation: dw = Vdy (9.9) where V is the quantity of liquid in the liquid stream. Substituting for dw in eqn. (9.8) we then have: Vdy/dt = Kl A(ys - y) which can then be integrated over time t during which time the concentration goes from an initial value of yo to a concentration y, giving loge [(ys - yo)/ (ys - y)] = tkl A/V. (9.10) Equation (9.10) shows, as might be expected, that the approach to equilibrium is exponential with time. The equation cannot often be applied because of the difficulty of knowing or

17 measuring the interfacial area A. In practice, suitable extraction times are generally arrived at by experimentation under the particular conditions that are anticipated for the plant. Stage Equilibrium Extraction Analysis of an extraction operation depends upon establishing the equilibrium and operating conditions. The equilibrium conditions are, in general, simple. Considering the extraction of a solute from a solid matrix, it is assumed that the whole of the solute is dissolved in the liquid in one stage, which in effect accomplishes the desired separation. However, it is not possible then to separate all of the liquid from the solid because some solution is retained with the solid matrix and this solution contains solute. As the solid retains solution with it, the content of solute in this retained solution must be then progressively reduced by stage contacts. For example, in the extraction of oil from soya beans using hexane or other hydrocarbon solvents, the solid beans matrix may retain its own weight, or more, of the solution after settling. This retained solution may therefore contain a substantial proportion of the oil. The equilibrium conditions are simple because the concentration of the oil is the same in the external solution that can be separated as it is in the solution that remains with the seed matrix. Consequently, y, the concentration of oil in the "light" liquid stream, is equal to x, the concentration of oil in the solution in the "heavy" stream accompanying the seed matrix. The equilibrium line is, therefore, plotted from the relation y = x. The operating conditions can be analysed by writing mass balances round the stages to give the eqn. (9.5). The plant is generally arranged in the form of a series of mixers, followed by settlers in which the two streams are separated prior to passing to the next stage of mixers. For most purposes of analysis, the solid matrix need not be considered; the solids can be thought of as just the means by which the two solution streams are separated after each stage. So long as the same quantity of solid material passes from stage to stage, and also the solids retain the same quantity of liquid after each settling operation, the analysis is straightforward. In eqn. (9.5), V refers to the liquid overflow stream from the settlers, and L to the mixture of solid and solution that is settled out and passes on with the underflow. If the underflow retains the same quantity of solution as it passes from stage to stage, eqn. (9.5) simplifies to eqn. (9.6). The extraction operation can then be analysed by application of step-by-step solution of the equations for each stage, or by the use of the McCabe-Thiele graphical method. EXAMPLE 9.6. Counter current extraction of oil from soya beans with hexane Oil is to be extracted from soya beans in a counter current, stage contact, extraction apparatus, using hexane. If the initial oil content of the beans is 18%, the final extract solution is to contain 40% of oil, and if 90% of the total oil is to be extracted, calculate the number of contact stages that are necessary. Assume that the oil is extracted from the beans in the first mixer, that equilibrium is reached in each stage, and that the crushed bean solids in the underflow retain in addition half their weight of solution after each settling stage. The extraction plant is illustrated diagrammatically in Fig. 9.5.

18 Figure 9.5 Hexane extraction of oil from soya beans in stages Each box represents a mixing/settling stage and the stages are numbered from the stage at which the crushed beans enter. The underflow will be constant from stage to stage (a constant proportion of solution is retained by the crushed beans) except for the first stage in which the entering crushed beans (bean matrix and oil) are accompanied by no solvent. After the first stage, the underflow is constant and so all stages but the first can be treated by the use of eqn. (9.6). To illustrate the principles involved, the problem will be worked out from stage-by-stage mass balances, and using the McCabe-Thiele graphical method. Basis for calculation: 100kg raw material (beans and their associated oil) entering stage 1. Concentrations of oil will be expressed as weight fractions. Overall mass balance In 100kg raw material there will be 18% oil, that is 82kg bean solids and 18kg oil. In the final underflow, 82 kg bean solids will retain 41 kg of solution, the solution will contain 10% of the initial oil in the beans, that is, 1.8kg so that there will be (18-1.8) = 16.2kg of oil in the final overflow, Extract contains (16.2 x 60/40) = 24.3kg of solvent Total volume of final overflow = = 40.5kg Total solvent entering = ( ) = 63.5kg Note that the solution passing as overflow between the stages is the same weight as the solvent entering the whole system, i.e. 63.5kg. MASS BALANCE Basis: 100kg beans Mass in (kg) Mass out (kg) Underflow Underflow Raw beans Extracted beans + solution Bean solids 82.0 Bean solids 82.0 Oil 18.0 Oil 1.8 Solvent 39.2 Overflow Overflow Solvent 63.5 Total extract 40.5 Solvent 24.3 Oil 16.2 Total Total 163.5

19 Analysis of stage 1 Oil concentration in underflow = product concentration = 0.4. It is an equilibrium stage, so oil concentration in underflow equals oil concentration in overflow. Let y2 represent the concentration of oil in the overflow from stage 2 passing in to stage 1. Then oil entering stage 1 equals oil leaving stage 1. Therefore balance on oil: 63.5y = 41 x x 0.4 y2 = Analysis of stage 2 x2 = y2 = 0.23 Therefore balance on oil: 41 x y3 = 63.5 x x 0.23 y3 = Analysis of stage 3 x3 = y3 = 0.12 Therefore balance on oil: 41 x y4 = 63.5 x x 0.12 y4= Analysis of stage 4 x4 = y4 = 0.049, Therefore balance on oil 41 x y5 = 63.5 x x 0.049, y5 = The required terminal condition is that the underflow from the final nth stage will have less than 1.8 kg of oil, that is, that xn is less than 1.8/41 = Since xn = yn and we have calculated that y5 is which is less than (whereas y4 = was not), four stages of contact will be sufficient for the requirements of the process. Using the graphical method, the general eqn. (9.6) can be applied to all stages after the first. From the calculations above for the first stage, we have x1 = 0.4, y2 = 0.23 and these can be considered as the entry conditions xa and ya for the series of subsequent stages. Applying eqn. (9.6): yn+1 = xn L/V + ya - xa L/V Now L = 41 V = 63.5 ya = 0.23 xa = 0.4 Therefore the operating line equation is: yn+1 = xn The equilibrium line is: yn = xn

20 The operating line and the equilibrium line have been plotted on Fig Figure 9.6 Hexane extraction of oil from soyabeans: McCabe - Thiele plot The McCabe -Thiele construction has been applied, starting with the entry conditions to stage 2 (stage 1 being the initial mixing stage) on the operating line, and it can be seen that three steps are not sufficient, but that four steps give more than the minimum separation required. Since the initial stage is included but not shown on the diagram, four stages are necessary, which is the same result as was obtained from the step-by-step calculations. The step construction on the McCabe -Thiele diagram can also be started from the nth stage, since we know that yn +1 which is the entering fresh solvent, equals 0. This will also give the same number of stages, but it will apparently show slightly different stage concentrations. The apparent discrepancy arises from the fact that in the overall balance, a final (given) concentration of oil in the overflow stream of was used, and both the step-by-step equations and the McCabe-Thiele operating line depend upon this. In fact, this concentration can never be reached using a whole number of steps under the conditions of the problem and to refine the calculation it would be necessary to use trial-and-error methods. However, the above method is a sufficiently close approximation for most purposes. In some practical extraction applications, the solids may retain different quantities of the solvent in some stages of the plant. For example, this might be due to rising concentrations of extract having higher viscosities. In this case, the operating line is not straight, but step-bystep methods can still be used. For some of the more complex situations other graphical methods using triangular diagrams can be employed and a discussion of their use may be found in Charm (1980), Coulson and Richardson (1978) or Treybal (1980).

21 It should be noted that in the chemical engineering literature what has here been called extraction is more often called leaching, the term extraction being reserved for liquid/liquid contacting using immiscible liquids. "Extraction" is, however, in quite general use in the food industry to describe processes such as the one in the above example, whereas the term "leaching" would probably only cause confusion. Washing Washing is almost identical to extraction, the main distinction being one of emphasis; in washing the inert material is the required product, and the solvent used is water which is cheap and readily available. Various washing situations are encountered and can be analysed. That to be considered is one in which a solid precipitate, the product, retains water which also contains residues of the mother liquor so that on drying without washing these residues will remain with the product. The washing is designed to remove them, and examples are butter, casein and cheese washing in the dairy industry. Calculations on counter current washing can be carried out using the same methods as discussed under extraction, working from the operating and equilibrium conditions. In washing, fresh water is often used for each stage and the calculations for this are also straightforward. In multiple washing, the water content of the material is xw (weight fraction) and a fraction of this, x, is impurity, and to this is added yxw of wash water, and after washing thoroughly, the material is allowed to drain. After draining it retains the same quantity, approximately, of water as before, xw. The residual yxw of wash liquid, now at equilibrium containing the same concentration of impurities as in the liquid remaining with the solid, runs to waste. Of course in situations in which water is scarce, counter current washing may be worthwhile. The impurity which was formerly contained in xw of water is now in a mass (xw + yxw); its concentration x has fallen by the ratio of these volumes, that is to x [xw/(xw + yxw)]. So the concentration, remaining with the solid, x1 is given by: after one washing: after two washings: x1 = x[xw /xw(1 + y)] = x[1/(1 + y)] x2 = x1[1/(1 + y)] = x[1/(1 + y)] 2 and so after n washings: xn = x[1/(1 + y)] n (9.11) If, on the other hand, the material is washed with the same total quantity of water as in the n washing stages, that is nyxw, but all in one stage, the impurity content will be: xn = x[1/(1 + ny)] (9.12) and it is clear that the multiple contact washing is very much more efficient in reducing the impurity content than is single contact washing, both using the same total quantity of water.

22 EXAMPLE 9.7. Washing of casein curd After precipitation and draining procedures, it is found that 100kg of fresh casein curd has a liquid content of 66% and this liquid contains 4.5% of lactose. The curd is washed three times with 194kg of fresh water each time. Calculate the residual lactose in the casein after drying. Also calculate the quantity of water that would have to be used in a single wash to attain the same lactose content in the curd as obtained after three washings. Assume perfect washing, and draining of curd to 66% of moisture each time. 100kg of curd contain 66kg solution. The 66kg of solution contain 4.5% that is 3 kg of lactose. In the first wash ( ) = 260kg of solution contain 3kg lactose. In 66 kg of solution lactose remaining there will be (66/260) x 3 = 0.76kg. After the second wash the lactose remaining will be (66/260) x 0.76 = 0.19kg After the third wash the lactose remaining will be (66/260) x 0.19 = 0.048kg Or, after three washings lactose remaining will be 3 x (66/260) 3 (Similar to the stage analysis.) = 0.048kg So, after washing and drying 0.048kg of lactose will remain with 34kg dry casein so that lactose content of the product = 0.048/34.05 = 0.14% and total wash water = 3 x 194 = 582kg To reduce the impurity content to kg in one wash would require x kg of water, where (3 x 66)/(x + 66) = 0.048kg x = 4060kg and so the total wash water = 4060kg Alternatively using eqns. (9.11) and (9.12) xn = x[1/(1 + y)] = 3[1/( /66)] 3 = xn = x[1/(ny + 1)] = 3[1/(ny + 1)] ny = Total wash water = nyxw = 61.5 x 66 = 4060kg. Extraction and Washing Equipment The first stage in an extraction process is generally mechanical grinding, in which the raw material is shredded, ground or pressed into suitably small pieces or flakes to give a large contact area for the extraction. In some instances, for example in sugar cane processing and in the extraction of vegetable oils, a substantial proportion of the desired products can be removed directly by expression at this stage and then the remaining solids are passed to the